Let H and K be normal subgroups of G, prove that H intersect K is also normal in G.

Proof:

Since H and K are normal in G, we have

Now, let , then c must retain the property of H and K since c is in both.

Thus, , implies

Therefore H intersect K is normal in G.

Is this right?

No it's not right. The problem is that you are confusing the equation with the equation where c is an element of .

Looking at H rather than for a moment, the equation means that the sets aH and Ha have the same elements. This does not mean that ah=ha for each element h of H. Instead, it means that, given h in H, ah=h'a for some element h' of H (which may be different from h).

Now coming back to the subgroup , it is not true that if then . What you can say is that and . You then need to show that . A similar argument will then show the reverse inequality .

Again, use my version of the equivalent condition for being normal. for all .

Now let it means . So by normality. Thus, . Thus, . Q.E.D.

My point is you should learn the way I do it. My approach certainly does not look as elegant as saying the left-right cosets. But it is much more efficient because I am using single elements only while you are working with an entire set.